Problem: Find $\dfrac{d}{dx}\left[\dfrac{\cos^2(x)}{x}\right]$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{-2\cos(x)\sin(x)-1}{x^2}$ (Choice B) B ${-2\cos(x)\sin(x)}$ (Choice C) C $\dfrac{\cos(x)\Bigl(2x-\cos(x)\Bigr)}{x^2}$ (Choice D) D $\dfrac{-\cos(x)\Bigl(2x\sin(x)+\cos(x)\Bigr)}{x^2}$
Answer: $\dfrac{\cos^2(x)}{x}$ is a quotient of a composite function and another function. Let... $u(x)=x^2$ $v(x)=\cos(x)$ $w(x)=x$... then $\dfrac{\cos^2(x)}{x}=\dfrac{u\Bigl(v(x)\Bigr)}{w(x)}$. To find $\dfrac{d}{dx}\left[ \dfrac{\cos^2(x)}{x}\right]$, we will need to use the quotient rule and the chain rule! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[\dfrac{u\Bigl(v(x)\Bigr)}{w(x)}\right] \\\\ &=\dfrac{\dfrac{d}{dx}\left[u\Bigl(v(x)\Bigr)\right]w(x)-u\Bigl(v(x)\Bigr)w'(x)}{[w(x)]^2} \gray{\text{Quotient rule}} \\\\ &=\dfrac{u'\Bigl(v(x)\Bigr)v'(x)w(x)-u\Bigl(v(x)\Bigr)w'(x)}{[w(x)]^2} \gray{\text{Chain rule}} \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=2x$ $v'(x)=-\sin(x)$ $w'(x)=1$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}\dfrac{{u'\Bigl(v(x)\Bigr)}v'(x)w(x)-{u\Bigl(v(x)\Bigr)}w'(x)}{[w(x)]^2} \\\\ &=\dfrac{{2\cos(x)}(-\sin(x))(x)-{\cos^2(x)}\cdot 1}{(x)^2} \\\\ &=\dfrac{-\cos(x)\Bigl(2x\sin(x)+\cos(x)\Bigr)}{x^2} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left[ \dfrac{\cos^2(x)}{x}\right]=\dfrac{-\cos(x)\Bigl(2x\sin(x)+\cos(x)\Bigr)}{x^2}$.